STAT 330 Lecture 20

Reading for Today's Lecture: 10.1.

Goals of Today's Lecture:

• Do an ANOVA by hand.
• Learn the geometry of an ANOVA table.

Today's notes

The ANOVA Table for a 1 way layout or an I sample problem

We generally record the arithmetic of our analysis in a table called an ANOVA table.

Sum of Mean

Source df Squares Square F P I-1 n-I

Total n-1

We can fill in all the rest from this much.

The sum of squares decomposition in one example

The data consist of blood coagulation times for 24 animals fed one of 4 different diets. Here is the data:

A B C D

62 63 68 56

60 67 66 62

63 71 71 60

59 64 67 61

65 68 63

66 68 64 63 59

We have , and .

Here is the ANOVA table.

Sum of Mean

Source df Squares Square F P

Diet I-1 228

Error n-I=20 112

Total n-1=23 340

Before I continue with the mathematics of the table I want to highlight the conclusion. There is overwhelming evidence that the diets lead to different mean blood coagulation times.

To fill in the table I needed to calculate various statistics such as

Similarly for diets B, C and D gives

and so on.

In the following I write the data in a table and decompose the table into a sum of several tables. The 4 columns of the table correspond to Diets A, B, C and D. Later in the course we will do matrix linear algebra and then want to think of stacking up these 24 values into a single column vector but the tables save space.

The sums of squares of the entries of each of these arrays are intervals for differences between the 4 population means. On the left hand side . This is the uncorrected total sum of squares. The first term on the right hand side gives . This term is sometimes put in ANOVA tables as the Sum of Squares due to the Grand Mean but it is usually subtracted from the total to produce the Total Sum of Squares we usually put at the bottom of the table and often called the Corrected (or Adjusted) Total Sum of Squares. In this case the corrected sum of squares is the squared length of the table

which is 340.

The second term on the right hand side of the equation has squared length (which is the Treatment Sum of Squares produced by SAS). The formula for this Sum of Squares is

but I want you to see that the formula is just the squared length of the vector of individual sample means minus the grand mean. The last vector of the decomposition is called the residual vector and has squared length . Corresponding to the decomposition of the total squared length of the data vector is a decomposition of its dimension, 24, into the dimensions of subspaces. For instance the grand mean is always a multiple of the single vector all of whose entries are 1; this describes a one dimensional space. The second vector, of deviations from a grand mean lies in the three dimensional subspace of tables which are constant in each column and have a total equal to 0. Similarly the vector of residuals lies in a 20 dimensional subspace - the set of all tables whose columns sum to 0. This decomposition of dimensions is the decomposition of degrees of freedom. So 24 = 1+3+20 and the degrees of freedom for treatment and error are 3 and 20 respectively. The vector whose squared length is the Corrected Total Sum of Squares lies in the 23 dimensional subspace of vectors whose entries sum to 1; this produces the 23 total degrees of freedom in the usual ANOVA table.

General Calculations

Here is the general formulation of this example. Write the data out as a big vector, say,

Now we write

Writing the equation for a particular component of this vector we have

Here are the Geometric facts:

1. Y, A, T, and R are vectors in Euclidean space of dimensions.
2. A, T and R are all perpendicular.
3. The ANOVA identity is

   

   The squared lengths of these 4 vectors are called the Total Sum of squares ( ), the Sum of Squares due to the Grand Mean ( ), the Sum of Squares due to Treatment ( ) and the Sum of Squares due to Error ( ).

4. Usually we move the grand mean to the other side and write

   

   and

   

   The quantity is called the Corrected Total Sum of Squares.

Here is a proof that A and T are perpendicular. The proofs are similar for and .

Recall that if

and

then x and y are perpendicular ( ) if

Then